Question

Butane (C₄H₁₀(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO₂ , Hf = –393.5 kJ/mol ) and water (H₂O, Hf = –241.82 kJ/mol) according to the equation below. What is the enthalpy of combustion (per mole) of C₄H₁₀ (g)?

Answer 1

The enthalpy of combustion per mole of butane (C4H10) is -2877.4 kJ/mol.

Step-by-step explanation:

The enthalpy of combustion per mole of butane (C4H10) can be calculated using the enthalpies of formation of carbon dioxide (CO2) and water (H2O), along with the balanced chemical equation for the combustion reaction.

The enthalpy change (∆H) for the reaction can be determined by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products. The balanced equation for the combustion of butane is: 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g).

Using the enthalpies of formation for CO2 (-393.5 kJ/mol) and H2O (-241.82 kJ/mol), and the enthalpy of formation of butane (-125.6 kJ/mol), we can calculate the enthalpy of combustion: ∆H = (8 × -393.5 kJ/mol) + (10 × -241.82 kJ/mol) - (2 × -125.6 kJ/mol) = -2877.4 kJ/mol. Therefore, the enthalpy of combustion per mole of butane is -2877.4 kJ/mol.